1.1 Synopsis

Stellar astrophyiscs – the success story of 20th century astronomy --
requires a synthesis of most of basic physics (thermodynamics, quantum
mechanics, and nuclear physics). It underlies nearly all of
astronomy, from reionisation to galaxy evolution, from interstellar
matter to planets, and from supernovae and planetary nebulae to white
dwarfs, neutron stars, and black holes.

In this course, we will review these successes (roughly first four
weeks) and then discuss current topics and remaining puzzles (in four
two-week series, detailed content depending on interest).

Course texts

The main book we will use is Stellar Structure and Evolution (KW;
Kippenhahn & Weigert for first, plus Weiss for second edition;
Springer-Verlag, 1990, 2012). Especially for those who did not take
undergraduate astrophysics, I strongly recommend An Introduction to Modern Astrophysics, by Carroll & Ostlie (2nd edition;
Addison-Wesley, 2006). This book introduces more empirical knowledge
(and jargon) assumed known in KW, and is used for the UofT
undergraduate courses AST 221 and AST 320 (I'll try to refer to
relevant notes from those classes; have a look at the parts on stars
from the AST 320 notes and mini problem sets).

Another way to think about ionisation, etc. (not discussed in class)

Consider a fixed volume V at a fixed temperature T (or,
equivalently, constant ρ and T). In thermal equilibrium, systems go
to their most probable state, i.e., one maximizes entropy,
\(S=k \log Z\), where \(Z\) is the partition function, a sum over all
possible states i, weighted by \(\exp(-E_i/kT)\). Usually, one can
split contributions, e.g., for non-interacting photons, ions, and
electrons, one has \(Z=Z_{\gamma}\times{}Z_{e}\times{}Z_{i}\) (and thus \(S=k\sum\log Z\)).

In the volume, for one particle at some momentum \(p\), the number of
phase space elements available is \((V/h^3)\times4\pi{}p^{2}dp\), with a
probability \(\exp(-\epsilon_p/kT)\). The total number of phase space elements
is thus \(\sim{}(V/h^3)p_{th}^3\), where \(p_{th}\) is some typical momentum
associated with the temperature. Doing the integral gives the
Maxwellian and \(p_{th}=\sqrt{2\pi{}mkT}\). Maybe more insightful is follow
Baierlein 2001AmJPh..69..423B and define a typical size,
\(\lambda_{th}\equiv{}h/p_{th}\), the ``thermal De Broglie'' wavelength. Then, the
number of possible states is simply \(V/\lambda_{th}^3\). For a set of N
identical particles, the contribution to the partition function is
thus \[Z_N=\frac{[g(V/\lambda_{th}^3)\exp(-\epsilon/kT)]^N}{N!},\] where \(g\) is the
number of internal states, the factorial \(N!\) ensures we do not
overcount states where two particles are swapped, and \(\epsilon\) is an energy
cost beyond thermal kinetic energy there may be for having this
particle.

Let's apply this to pair creation, assuming some mix of photons, ions,
electrons and electron-positron pairs. Assuming a dilute plasma,
their contributions to \(Z\) can be split, i.e.,
\(Z=Z_{\gamma}\times{}Z_{e}\times{}Z_{i}\times{}Z_{\pm}\) (of course, the physical
picture is that there is a formation rate from the interactions of two
photons, balanced by an annihilation rate; for the statistics, we are
only concerned about the final equilibrium ). Since the electrons and
positrons are independent, \(Z_{\pm}=Z_{+}\times{}Z_{-}\), with both given by the
above equation with \(\epsilon=m_{e}c^2\), but with \(N_{+}=N_{-}=N_{\pm}\). Hence, \(Z_{+}=Z_{-}\),
and to find the number of particles, we can just find the maximum of
\(S_{+}=k\log{}Z_{+}\), i.e., \[\frac{\partial{}S_{+}}{\partial{}N_{+}} = \frac{\partial{}k\log Z_{+}}{\partial{}N_{+}} =
\frac{\partial}{\partial{}N_{+}}kN_{+}\left[\log\left(g\frac{V}{\lambda_{th}^3}\right)-\frac{m_{e}c^2}{kT}
-\log N_{+}-1\right]=0,\] where we used that for large \(N\), \(N!=N\log
N - N\). Solving this for \(N_{+}\), one finds \[
N=g\frac{V}{\lambda_{th}^3}\exp(-m_{e}c^2/kT).\] Equivalently, one has
\(n\equiv{}N/V=g\exp(-m_{e}c^2/kT)/\lambda_{th}^3\), which has the nice implication that for
classical particles, the probability for one with given internal state
to exist in a given volume element \(\lambda_{th}^3\) is simple \(\exp(-\epsilon/kT)\).
Thus, for this very small volume, the probability becomes significant
for \(kT\approx{}m_{e}c^2\). But when does the number of pairs become significant
on larger scales? One measure to use is when \(n_{\pm}=n_e\), i.e., when
\(\exp(-m_{e}c^2/kT)=_{}\lambda_{th}^3 n_e/g\). For electrons (\(m=m_e\)), one has
\(\lambda_{th}=2.4\times10^{-10}T_9^{-1/2}\) cm, and
\(n_e=\rho/\mu_e{}m_H=6\times10^{23}(\rho_2/\mu_e)\) cm-3, so it requires
\(T_9\approx{}m_{e}c^{2}/k(11.7+\log{}gT_9^{1/2}/\rho_2)\approx0.6\), quite
consistent with KW, Fig. 34.1.

One can treat ionisation similarly, writing
\(Z_H=Z_{0}\times{}Z_{p}\times{}Z_{e}\). We need to use that \(N_{p}=N_{e}=N_{H}-N_0\).
Doing a similar derivations as above, one derives the Saha equation.
Again, ionisation is well before \(kT\approx{}\chi\). One consequence of this, is
that if one, e.g., wants to know the population in excited states in
hydrogen, it is easier to do this relative to the ionised state (since
by the time you can excite even to the first excited state with
\(\epsilon_2=\chi_H(1-1/4)=10.2\) eV, hydrogen is mostly ionised). For given
state \(s\), one thus writes \(n(H_0,s)/n_{p} =
(g_s/g_{p}g_{e}n_{e}\lambda_{th}^3)\times\exp((\chi-\epsilon_2)/kT)\).

Finally, back to the chemical potential \(\mu\) (and Baierlein
2001AmJPh..69..423B). In terms of above quantities, one finds
\(\mu=\epsilon+kT\log(g\lambda_{th}^3/n)\), but \(\mu\) also enters all thermodynamic
potentials (internal energy U, enthalpy H, Helmholtz free energy F,
Gibbs free energy G), as an additional term \(\dots+\mu{}dN\), i.e., the energy
required to add one particle. In particular, for constant T, V,
Helmholtz is handiest: \(F(T,V,N)=PV+\sum_{i}\mu_{i}N_{i}\) (and
\(dF=PdV+\sum_{i}\mu_{i}dN_i\)). For pair plasma, minimizing \(F\) for \(N_{+}=N_{-}\)
(holding \(T\), \(V\), other \(N\) constant), one requires \(\mu_{+}+\mu_{-}=0\).
With the above microscopic definition of \(\mu\), one recovers the
solution. Similarly, for ionisation, \(\mu_0=\mu_p+\mu_e\). In general, for
any reaction left↔right, one expects that in equilibrium,
\(\sum_{left}\mu=\sum_{right}\mu\). (In that sense, the above are missing photons
– but these have \(\mu_\gamma=0\).)

All the above was for classical particles, but the same holds for
non-classical ones (except of course that one cannot assume a
Maxwellian once particles start to overlap, \(\lambda_{th}\approx{}d=n^{-1/3}\)).
For completely degenerate neutron gas, where \(\mu=\epsilon_F\), one now trivially
finds that there will be a contribution of protons and electrons such
that \(\mu_n=\mu_p+\mu_e\). (Here, there is no \(\mu_\nu\), since the neutrinos
escape; for a hot proto-neutron star, where the neutrino opacity is
still high, one does need to include it.) Remember, however, that
above we derive a final, equilibrium state. The process to get there
can be slow – not all baryons are in the form of iron yet!

Rosseland mean: 1/〈κ〉 = (π/acT3)∫ν(1/κν)(dBν/dT)dν

Criterion for convection: -(1/γ)dlnP/dr > dlnρ/dr

With composition gradients → ∇ad<∇rad-f∇μ,
where ∇μ=dlnμ/dlnP and
f=(∂lnρ/∂lnμ)/(-∂lnρ/∂lnT); f=1 for a fully-ionised ideal gas.

4.2 Wed Jan 30

Textbook

KW 6.2–6.5, 7, 30.4, 24 (AST 320 notes 6, 7)

Exercises

AST320 mini-PS V

Scalings for conduction

Generally, the flux is \(F=-\frac13vl\nabla{}U\). It can be separated in
different components. For photons, we saw \(U=aT^4\), \(v=c\) and \(l=1/\sigma{}n\)
and hence one has \(F=-(4ac/3)(T^3/\sigma{}n)\nabla{}T\) (where usually
we write \(\sigma{}n=\kappa\rho\), but it is easier not to do so here). Given the
definition of conductivity through \(F=-k\nabla{}T\), one infers an equivalent
conductivity \(k_{\gamma}=(4ac/3)(T^3/\sigma{}n)\).

For particles, \(U=\frac32nk_{B}T\) and thus
\(F=-\frac13vln\frac32k_{B}\nabla{}T\). Again writing \(l=1/n\sigma\), one finds
\(k=\frac13\frac23k_{B}(v/\sigma)\). For an ideal, completely ionised gas,
\(v\propto{}T^{1/2}\) and \(\sigma\sim{}Z^{2}e^4/(kT)^{2}\propto1/T^2\). Hence, \(k\propto{}T^{5/2}\).

For degenerate material, we should consider ions and electrons
separately. The ions still have very short mean-free path, so do not
contribute much. For the electrons, only a small fraction \(kT/E_{F}\)
near the Fermi surface carries any heat, i.e., \(U_{e}\sim{}n_{e}(kT/E_F)kT\),
and thus \(\nabla{}U\sim{}n_{e}(k_{B}T/E_{F})\nabla{}T\). Furthermore, those electrons have
velocity depending on density, not temperature. Their free path still
is \(l=1/n_{i}\sigma\) (\(n_i\) the ion density), but now
\(\sigma\sim{}Z^{2}e^4/E_{F}^2\propto{}1/E_{F}^2\), and thus
\(k_{e}\propto{}(v/\sigma{}n_{i})n_{e}(k_{B}T/E_{F})\propto{}vE_{F}T\). For non-relativistic electrons,
\(v\propto{}\rho^{1/3}\) and \(E_{F}\propto\rho^{2/3}\), so \(k_{e}\propto\rho{}T\). For relativistic
particles, \(v\to{}c\) and \(E_{F}\propto\rho^{1/3}\), so \(k_{e}\propto\rho^{1/3}T\).

Writing in terms of an equivalent opacity, \(\kappa=(4ac/3)(T^3/k\rho)\), one
finds for the ionised ideal gas, the opacity for electrons scales as
\(\kappa_{e}\propto{}T^{1/2}/\rho\), for non-relativistic degenerate electrons,
\(\kappa_{e}\propto{}T^2/\rho^2\), and for relativistic degenerate electrons,
\(\kappa_{e}\propto{}T^2/\rho^{4/3}\) (the former two are as mentioned by Elliot).
Note that the photon opacity should also be affected, since photons
can only interact with electrons near the Fermi surface, so
\(l_{\gamma}\sim1/\sigma{}n_{e}(kT/E_{F})\). Equivalently, one can write that
the effective opacity scales as \(T/E_{F}\propto{}T/\rho^{2/3}\)
(non-relativistic) or \(T/\rho^{1/3}\) (relativistic). At high densities,
however, electron conduction will still win.

5.2 Wed Feb 6 (only first hour)

Short presentation

5.3 Mon Feb 11

Textbook

KW 18 (except 18.5.3, beyond helium)

Exercises

AST320 mini-PS XI, XII

Temperature dependence

Generally, we write the cross section
\(\sigma(E)=(S(E)/E)\exp(-b/\sqrt{E})\), and integrate over \(E\) to get
\(\langle\sigma{}v\rangle\), i.e.,
\(
\langle\sigma{}v\rangle=\sqrt{\frac{8}{\pi\mu}}\left(\frac{1}{kT}\right)^{3/2}
\int_E S(E)\exp(-E/kT-b/\sqrt{E}) dE
\)
We discussed how normally \(S(E)\) can be taken out of the integral and
one finds the Gamov peak, with height \(\exp(-3E_0/kT)\), with
\(3E_0/kT=-19.721(\mu/m_u)^{1/3}(Z_{a}Z_{b})^{2/3}T_{7}^{-1/3}\).

As Charles mentioned, resonances are important. The above holds if
one's energy is in the far wing of a resonance, so that \(S(E)\) indeed
varies slowly. But if the resonance is inside the Gamov peak, it can
dominate the energy dependence. In that case, one can consider it as
a delta function, and the reaction rate will scale just with
\(\exp(-E_{res}/kT-b/\sqrt{E_{res}})\), i.e., the only temperature-dependent
part comes from how many particles have the right energy. For this
reason, the \(3\alpha\) reaction rate has a term with \(\exp(-C/T)\) instead of
\(\exp(-C/T^{1/3})\).